# Distinguished triangle in abelian triangulated categories

I know that an abelian triangulated category is semisimple,i.e.,any exact sequence splits.But Why does any distinguished triangle is isomorphic to a triangle of the form $$X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} \operatorname{ker} f[1] \oplus \operatorname{coker} f \stackrel{h}{\rightarrow} X[1]$$?(cf. Gelfand and Manin, Exercises to IV 1).

My try:Let $$X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} Z \stackrel{h}{\rightarrow} X[1]$$ be a distinguished triangle.Then we have $$\operatorname{coker} f \stackrel{g}{\rightarrow} Z \stackrel{h}{\rightarrow} \operatorname{ker} f[1]$$.If it is exact at $$\operatorname{coker} f,Z$$,and $$\operatorname{ker} f[1]$$,then we have $$Z\cong \operatorname{ker} f[1] \oplus \operatorname{coker} f$$.But I can't show the exactness.

• If the category is semisimple, then split the map $f:X\to Y$ so that you have $X=Ker(f)\oplus X'$ and $Y=Coker (f)\oplus Y'$. Noe you can describe your map $f$ as being 0 on its kernel and an iso from $X'$ into Coker(f). The result now folllows from the fact that direct suma of tengles are rriangle – Marco Farinati Jun 12 at 23:06

Split $$X=Ker(f)\oplus X'$$, and $$Y\cong X'\oplus Coker(f)$$. Now you can describe $$f$$ as a direct sum of 3 maps, zero in $$Ker(f)$$, identity on $$X'$$ and $$0\to Coker(f)$$. The result follows from the fact that identity fits into a triangle with 0 as cone, and the zero maps similarly (just traslate the triangle with identity)